Determinant

Determinant#

The determinant is a scalar quantity associated with any square matrix $A \in R^{n \times n}$ . It encodes important geometric and algebraic information about the transformation represented by $A$ .

Geometrically, the determinant tells us:

How the matrix $A$ scales volume: The absolute value $| det (A) |$ is the volume-scaling factor for the linear transformation $x \mapsto A x$ .
Whether the transformation preserves or flips orientation: If $det (A) > 0$ , the transformation preserves orientation; if $det (A) < 0$ , it reverses it (like a reflection).

Algebraically, the determinant can be defined as:

det (A) = \sum_{σ \in S_{n}} sgn (σ) \cdot a_{1 σ (1)} a_{2 σ (2)} \dots a_{n σ (n)}

where:

The sum is over all permutations $σ$ of ${1, 2, \dots, n}$ ,
$sgn (σ)$ is $+ 1$ or $- 1$ depending on the parity of the permutation.

This formula is computationally expensive and confusing, but conceptually important: it captures how the determinant depends on all possible signed products of entries, each taken once from a distinct row and column.

Let’s illustrate the determinant geometrically.

../_images/ee97e66adb3b31b886ee6c5d93bddacaf3acd491650ebc58f47abb5acfa69ec0.png

Identity: No change — area = 1.
Stretch: Expands area — determinant > 1.
Flip: Reflects across the diagonal — determinant < 0.
Rotation: Rotates without distortion — determinant = 1.

What is the Determinant?#

The determinant of a matrix $A \in R^{n \times n}$ is a scalar that describes how $A$ scales space. Algebraically, it is defined by a signed sum over all permutations of the matrix’s entries. Geometrically, it quantifies the change in signed volume of a unit shape under transformation by $A$ . If the determinant is zero, the transformation collapses the volume entirely, and $A$ is singular (non-invertible).

The determinant has several important properties:

(i) $det (I) = 1$

(ii) $det (A^{⊤}) = det (A)$

(iii) $det (A B) = det (A) det (B)$

(iv) $det (A^{- 1}) = det (A)^{- 1}$

(v) $det (α A) = α^{n} det (A)$

Practical Computation of the Determinant#

The algebraic definition of the determinant is computationally expensive. In practice, we compute the determinant using property (iii) and a matrix factorization such as the PLU decomposition:

A = P L U

where $P$ is a permutation matrix, $L$ is a unit lower triangular matrix, and $U$ is an upper triangular matrix.

Theorem (Triangular Matrix Determinant)

Let $T \in R^{n \times n}$ be a triangular matrix, either upper or lower triangular.

Then:

det (T) = \prod_{i = 1}^{n} T_{i i}

Then,

det (A) = det (P) \cdot det (L) \cdot det (U)

Since:

$det (L) = 1$ (if unit lower triangular),
$det (U) = \prod_{i = 1}^{n} u_{i i}$ ,
$det (P) = (- 1)^{s}$ , where $s$ is the number of row swaps,

this method reduces determinant computation to $O (n)$ operations after LU decomposition. As the cost for the LU decomposition is $O (n^{3}),$ the total cost of computing the determinant is $O (n^{3}) .$

Cofactor Expansion: Definition#

Cofactor expansion (also called Laplace expansion) gives a recursive definition of the determinant.

Let $A \in R^{n \times n}$ be a square matrix.

Then the determinant of $A$ can be computed by expanding along any row or column.

For simplicity, we’ll define it for the first column.

det (A) = \sum_{i = 1}^{n} (- 1)^{i + 1} \cdot A_{i 1} \cdot det (A^{(i, 1)})

Where:

$A_{i 1}$ is the entry in row $i$ , column 1
$A^{(i, 1)}$ is the minor matrix obtained by deleting row $i$ and column 1 from $A$
$(- 1)^{i + 1}$ is the sign factor for alternating signs (from the checkerboard sign pattern)
$(- 1)^{i + j} \cdot det (A^{(i, j)})$ is called the cofactor of $A_{i j}$

This formula recursively reduces the computation of a determinant to smaller and smaller submatrices.

Cofactor Expantion Example (3×3 Matrix)#

Let:

\begin{array}{r} A = [\begin{array}{c} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}] \end{array}

Expand along the first column:

\begin{array}{r} det (A) = (+ 1) \cdot 1 \cdot | \begin{array}{c} 5 & 6 \\ 8 & 9 \end{array} | - 4 \cdot | \begin{array}{c} 2 & 3 \\ 8 & 9 \end{array} | + 7 \cdot | \begin{array}{c} 2 & 3 \\ 5 & 6 \end{array} | \end{array}

Now compute the 2×2 determinants:

det (A) = 1 \cdot (5 \cdot 9 - 6 \cdot 8) - 4 \cdot (2 \cdot 9 - 3 \cdot 8) + 7 \cdot (2 \cdot 6 - 3 \cdot 5)

= 1 \cdot (- 3) - 4 \cdot (- 6) + 7 \cdot (- 3) = - 3 + 24 - 21 = 0

So:

det (A) = 0

Proof. via Laplace Expansion / Cofactor Expansion

We’ll prove this for upper triangular matrices by induction on the matrix size $n$ . The same argument applies symmetrically for lower triangular matrices.

Base Case: $n = 1$

Let $T = [t_{11}]$ . Then clearly:

det (T) = t_{11} = \prod_{i = 1}^{1} T_{i i}

The base case holds.

Inductive Step

Assume the result holds for $(n - 1) \times (n - 1)$ upper triangular matrices.

Now let $T \in R^{n \times n}$ be upper triangular. That means all entries below the diagonal are zero:

\begin{array}{r} T = [\begin{array}{c} t_{11} & t_{12} & \dots & t_{1 n} \\ 0 & t_{22} & \dots & t_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & \dots & t_{n n} \end{array}] \end{array}

Use cofactor expansion along the first column. Since the only nonzero entry in the first column is $t_{11}$ , we have:

det (T) = t_{11} \cdot det (T^{(1, 1)})

Where $T^{(1, 1)}$ is the $(n - 1) \times (n - 1)$ matrix obtained by deleting row 1 and column 1. But:

$T^{(1, 1)}$ is still upper triangular.
By the inductive hypothesis:

$det (T^{(1, 1)}) = \prod_{i = 2}^{n} T_{i i}$

So:

det (T) = t_{11} \cdot \prod_{i = 2}^{n} T_{i i} = \prod_{i = 1}^{n} T_{i i}

The inductive step holds.

Conclusion

By induction, for any upper (or lower) triangular matrix $T \in R^{n \times n}$ ,

det (T) = \prod_{i = 1}^{n} T_{i i}

The determinant accumulates only the diagonal entries because each pivot is isolated, and all other paths in the expansion have zero entries.
This result is frequently used in:
- Computing determinants from LU decomposition
- Checking invertibility efficiently
- Proving properties of eigenvalues and characteristic polynomials